3t(t+5)=-18

Solution for 3t(t+5)=-18 equation:

3t(t+5)=-18

We move all terms to the left:

3t(t+5)-(-18)=0

We add all the numbers together, and all the variables

3t(t+5)+18=0

We multiply parentheses

3t^2+15t+18=0

a = 3; b = 15; c = +18;
Δ = b2-4ac
Δ = 152-4·3·18
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-3}{2*3}=\frac{-18}{6}
=-3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+3}{2*3}=\frac{-12}{6}
=-2
$